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Can someone help me with this question please? (Chemistry question from BMAT paper) Watch

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    Thank you in advance!
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    Have you tried doing this question? Remember hydrogen gas is diatomic.
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    (Original post by PleaseHelppppp)
    I think its D.


    Moles of sodium = 1.15/23 = 0.05

    I believe its a one to one ratio?

    Volume = mol x 22.4 so 0.05 x 22.4 is 1.12 dm3. To convert 1.12 dm3 into cm3, times by 1000 so 1.12 x 1000 gives you 1120.
    This is incorrect, please see my post.
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    It's B, 560cm^3.

    Alright, so firstly you have to write out the balanced equation for this reaction.
    2Na + 2H2O -----> 2NaOH + H2

    Now, you need to calculate the molar mass of Na which is given to be 23 g/mol. The mass is also given for Na as 1.15g. Using the equation mass / molar mass = moles you can then calculate the moles of Na in the equation. 1.15 / 23 = 0.05 mol.

    Now you know the moles of Na is 0.05 you can then work out the moles for what you're trying to calculate, which is hydrogen gas.
    Since the ratio of Na to H2 is 2:1 (you can see this by looking at the coefficient of Na, being 2 and the coefficient of H2 being 1) this means that you have to divide the moles you calculated from Na (which was 0.05 mol) by 2. So 0.05/2 is 0.025 mol of H2.

    Now you know the moles of hydrogen gas you can implement it into the formula for calculating volume. moles * molar volume = volume. The question tells us that at STP the volume is 22.4dm^3 so that would be the molar volume. However, to convert dm^3 to cm^3 you have to multiply by 1000 so the molar volume is 22400cm^3. Implement this into the formula moles * molar volume = volume and it is 0.025 * 224000 = 560. So your answer is 560cm^3.
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    (Original post by Zoz_k)
    How do you know how to write out an equation? I'm alright with balancing but I have no idea how to write out an equation.
    Ok so the question says that sodium (Na) reacts with water (H2O) so you know that those are the reactants. The question also says that hydrogen is one of the products and we know that hydrogen comes in pairs as it's a diatomic molecule so we know one of the products is H2. Ultimately you need to know that when an alkali metal reacts with water it produces a metal hydroxide (in this case NaOH) and it also produces hydrogen (which is given). There are general rules that you just have to remember, but in some cases you can work out the reactants by rearranging the products. In this case, you need to have just remembered the rule that when an alkali metal reacts with water it produces a metal hydroxide and hydrogen. This was highlighted in GCSE Chemistry so if you have forgotten some of these rules then you may need to go over them.
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    (Original post by Deliciate)
    It's B, 560cm^3.

    Alright, so firstly you have to write out the balanced equation for this reaction.
    2Na + 2H2O -----> 2NaOH + H2

    Now, you need to calculate the molar mass of Na which is given to be 23 g/mol. The mass is also given for Na as 1.15g. Using the equation mass / molar mass = moles you can then calculate the moles of Na in the equation. 1.15 / 23 = 0.05 mol.

    Now you know the moles of Na is 0.05 you can then work out the moles for what you're trying to calculate, which is hydrogen gas.
    Since the ratio of Na to H2 is 2:1 (you can see this by looking at the coefficient of Na, being 2 and the coefficient of H2 being 1) this means that you have to divide the moles you calculated from Na (which was 0.05 mol) by 2. So 0.05/2 is 0.025 mol of H2.

    Now you know the moles of hydrogen gas you can implement it into the formula for calculating volume. moles * molar volume = volume. The question tells us that at STP the volume is 22.4dm^3 so that would be the molar volume. However, to convert dm^3 to cm^3 you have to multiply by 1000 so the molar volume is 22400cm^3. Implement this into the formula moles * molar volume = volume and it is 0.025 * 224000 = 560. So your answer is 560cm^3.

    Thank you so much!!
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    (Original post by Wacky01)
    Thank you so much!!
    np, glad I could help
 
 
 
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